SOLUTION: Hello, I have an assignment I've been scratching my head about. Person A sends messages every 16 minutes, Person B - every 16 minutes, Person C - every 14 minutes. What is the p

Algebra ->  Probability-and-statistics -> SOLUTION: Hello, I have an assignment I've been scratching my head about. Person A sends messages every 16 minutes, Person B - every 16 minutes, Person C - every 14 minutes. What is the p      Log On


   

Question 1191120: Hello, I have an assignment I've been scratching my head about.
Person A sends messages every 16 minutes, Person B - every 16 minutes, Person C - every 14 minutes. What is the probability of receiving at least ONE message in the span of 6 minutes? JUST ONE message in the span of 6 minutes? And JUST TWO messages in the span of 6 minutes?
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


From the given conditions....

P(A sends a message in a span of 6 minutes) = 6/16 = 3/8
P(A does not send a message in a span of 6 minutes) = 5/8
P(B sends a message in a span of 6 minutes) = 6/16 = 3/8
P(B does not send a message in a span of 6 minutes) = 5/8
P(C sends a message in a span of 6 minutes) = 6/14 = 3/7
P(A does not send a message in a span of 6 minutes) = 4/7

Then....

P(only A sends a message in a span of 6 minutes) = (3/8)(5/8)(4/7) = 60/448
P(only B sends a message in a span of 6 minutes) = (5/8)(3/8)(4/7) = 60/448
P(only C sends a message in a span of 6 minutes) = (5/8)(5/8)(3/7) = 75/448

ANSWER #1: P(receiving only 1 message in a span of 6 minutes) = 195/448

P(A and B send a message in a span of 6 minutes) = (3/8)(3/8)(4/7) = 36/448
P(A and C send a message in a span of 6 minutes) = (3/8)(5/8)(3/7) = 45/448
P(B and D send a message in a span of 6 minutes) = (5/8)(3/8)(3/7) = 45/448

ANSWER #2: P(receiving two messages in a span of 6 minutes) = 126/448

Simplify the fraction answers or convert to decimals or percentages as required/desired.